Chapter 3 Rolling Motion

Mr. Faisal Ikram bin Abd Samad

Equation of rolling motion can written as follows:

tcosMcdtd

bdt

da eo2

2

=+

+

z

Rolling

x

y

Equation of rolling motion in calm water is

or

This equation can also be written as

Where and

0'IGMg

dtd

'Ib

dt

d

xx

T

xx2

=

+

+2

02 2 =++

xx'Ib

ab

2 xx

T2

'IGMg

ac

0GMgdtd

bdt

d'I T2xx =+

+

2

Here is the natural frequency (circular) for rolling. The expression for the natural rolling period, T is given as

Now, the solution is written as

or

Where

2T

)tsinAtcosA(e d2d1t +

)tsin(Ae dt

22d

Example Calculation – 2.14 L = 150 mKxx = 0.40 B GMT = 2.5 mB = 20 m = 15,000 tonnes If the added mass moment of inertia is about 90 %, find the natural rolling period !

Example Calculation – 2.14

2

22

mtonnes500,712

)204.0(000,1590.1K90.1axx

=

==

kNm875,3675.2x81.9x000,15GMgc T ===

sec/rad719.0500,712875,367

ac

===

sec734.82

T =

=

Equation of rolling motion in waves is

The steady state solution is usually taken for analysis.

= a sin (e.t -2) or

=

tsinMGMgdtd

bdt

d'I e0T2xx =+

+

2

)t(sink4)1(

2e2222

st -

The amplitude of the forced rolling motion a is given by: a = st.

Where, st = static rolling amplitude =

= magnification factor =

=

= tuning factor = =

cMo

st

a

.freq.Natencounterof.Freq

2222 k4)1(

1

e

= non-dimensional damping factor =

= and 2 = phase angle between the exciting moment and the motion = tan –1

21

k2

2bI'xx

c=

I'xx


Given :L = 140m Cwp = 0.80 B = 19mGMT = 5m T = 8.5m Kyy = 32mCB = 0.65 = 30o V = 15knotsa = 1.5m = 1025 kg/m3

i) Natural roll period of ship is 6.885 seconds

ii) Non-dimensional damping factor is 0.125


iii) The non-dimensional amplitude of the rolling moment,

fo =

(Where, w = L and Cwp = 0.8)

Find :a) Amplitude of the rolling motionb) Phase differences between the wave and the

rolling motion

25.0TLBgk

M2

a

o =


Restoring moment coefficient,

c = .g.GMT = 15063913 X 9.81 X 5 kg.m.m/s2

= 3.694 x 108 N.m/rad.

Natural frequency for rolling,

=

= 0.912 rad/sec.

885.62

T2


The wave number is

Tuning factor, = e/ = 0.363/0.912 = 0.398

Static roll deflection, st =

= 0.1973 rad. (11.31o)

The amplitude ratio, =

= 1.18

1

W

m045.0140

28.6L2

k -==

=

cmo

2222st

a

4)1(

1

K+-=


Amplitude of roll motion, a = x st = 1.18 x 0.1973

= 0.233 rad. (13.35o)

The phase angle bet. wave motion and roll motion,

= 1 + 2

2 = tan-1 = tan-1

= 6.74o

= - 90 + 6.74 = - 83.26o

21

.k.2

- 2398.01398.0x125.0x2


wave excitation moment

rolling motion

A

B

CA

B

C

e1

e2

e1

e2

w

t

a = Ixx + Ixx

Can also be written as:a = (m + m) x kxx

2

For normal shipskxx = 0.33 B 0.45 B (B = ship breadth)

Usually approximated as 10 to 20% of the actual ship mass. This value is used when there is no available data on added mass.

Damping forces arises from the combination of any of the following: a) Waves generatedb) Water friction on the ship surface or eddy makingc) Bilge keels, skeg, and other appendagesd) Resistance between the ship and the aire) Energy loss because of heat generated during the rolling motionf) Surface tension

Damping coeff per unit length,

bn =

Where Ā = = d

d is a function of individual section coefficient and Sn.

Sn = Bn/2Tn

-

22n

3e

2

A2

Bg

)2B( n

a

a

2

n2

e

g2B

Example Calculation - 2.16 Find the damping coefficient for rolling for the model described in Example Calculation – 2.5.

StationNo

Bn Tn Bn2 S

n

1 2 3 4 5 6 705101520

00.7900.7900.790

0

0.3490.3490.3490.3490.349

0.00.9320.9320.932

0.0

00.62410.62410.6241

0

0.00.1560.1560.156

0.0

0.00.27350.27350.27350.

0

n

2

e Bxg2

w

4B2

n

Example Calculation - 2.16

Damping coefficient for rolling= bn d = 1/3 x station spacing x product= 1/3 x 1.463 x 2825.4 = 1,377.85 Nms

n=nn

nTB

S df

(4)2 x 9

bn

SM

Product

8 9 10 11 12 0.0

0.992 0.992 0.992 0.0

0.0 0.76 0.76 0.76 0.0

0 0.66 0.66 0.66

0

0 0.4356 0.4356 0.4356

0

0 282.54 282.54 282.54

0

1 4 2 4 1

0 1130.16

565.08 1130.16

0

SUM 2825.4

A 2A

116g3e

2

Damping coefficient can also be obtained from experiment in calm water.

The damping ratio = b/bc bc = 2.I’xx.wn

t

Imaxl = o e-nt

t1 t2 t3

0

10

5

- 5

- 10

15

- 15

(Deg)

The ratio of two amplitudes, at times t1 and t3 (over a period of time T),

where &

By obtaining value of from experiment, we can now calculate value of

b = bc. n = rad/sec. = 2.I’xx.n.

3

1

xxI'c

3

1

21

2

=ln 2

d nd

2

T

Restoring moment = B.GZ (for small angle of inclination) Thus, c. = B.GZ

= B.GMT.sin

B.GMT.

c = B.GMT or .g..GMT

G z

. g

Bf

fm

i.e f < 100

f z

y

P

RQ

2/3 y

The amplitude is,

Mo =

and 1 = -90o since sin = cos ( – /2 )

The non dimensional amplitude of the exciting moment is

fo =

- 2/L

2/L

3a dx.y)cos.x.kcos(sin..k.g.

32

= TBLkg

M2

a

0 - 2/L

2/L

32

dx.y)cos.x.kcos(TBL

sin32

Example Calculation – 2.17 Using the data from Example Calculation – 2.7, calculate the exciting moment for the rolling motion.

StationNo.

y = B n /2 (m) y3

(m3) cosk

cos(k cos)

y3

cos(k cos

) SM Product

2 3 4 5 6 7 8 9

05

101520

00.40.40.40

- 2.93- 1.465

01.4652.93

00.0640.0640.064

0

1.5700.7850.000

- 0.785- 1.570

0.0000.7071.0000.7070.000

0.0000.04520.0640.04520.000

14241

0.0000.18080.1280.18080.000

SUM 0.4896

Example Calculation – 2.13 Integral = 1/3 x station spacing x SUM

= 1/3 x 1.465 x –3.312 = -1.617 m3

Therefore

or

235.0617.1xLB

4egralintx

LB

4f

220 -=-==

Nm95.899,1fLBg21

M 02

a0 ==

Example Calculation – 2.13 Integral = 1/3 x station spacing x SUM

= 1/3 x 1.465 x 0.4896 = 0.239 m3

Therefore

or

105.0239.0xTBL

866.032

egralintxTBL

sin32

f220 ==

=

Nm80.86fTBLkgM 02

a0 ==

Chapter 3 Rolling Motion

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